Characterization of those semigroups with zero on which every equivalence relation is a congruence

Let S be a semigroup with zero. Show that if every equivalence relation on S is a congruence, then S is either a zero semigroup or a semilattice of order 2.


Let S be a semigroup with a zero element 0, and suppose that every equivalence relation on S is a congruence.

Suppose a,b \in S such that ab \neq 0. Consider now the equivalence relation \sigma on S whose classes are \{\{a,0\}, S \setminus \{a,0\}\}. By our hypothesis, \sigma is a congruence. Now a \sigma 0, so ab \sigma 0b = 0. Since ab \neq 0, we have ab = a. Now consider the equivalence \tau on S whose classes are \{\{b,0\}, S \setminus \{b,0\}\}. Again \tau is a congruence, and we have b \tau 0, so that ab \tau a0 = 0. Since ab \neq 0, we have ab = b. So a = b. Moreover, a^2 = a.

Thus we have S = Z \cup E, where Z = \{s \in S \ |\ sS = 0\} and E = S \setminus Z. By the preceding argument, in fact E = \{s \in S\ |\ s^2 = s, s \neq 0\} is the set of nonzero idempotents in S.

We claim that E contains at most one element. To see this, suppose E \neq \empty and let e,f \in E. Consider the equivalence \sigma on S whose classes are \{\{e,f\},S \setminus \{e,f\}\}. Now e \sigma f, so that e = ee \sigma ef. So either ef = e or ef = f. If ef = e, consider the equivalence \tau whose classes are \{\{f,0\}, S\setminus\{f,0\}\}. Now f \tau 0, so ef \tau 0. Since ef = e \neq 0, ef = f, and thus e = f. If ef = f, consider the equivalence \tau whose classes are \{\{e,0\}, S \setminus \{e,0\}\}. By a similar argument, e = f. So E contains at most one element.

If E = \emptyset, then S = Z is a zero semigroup.

Suppose instead that E = \{e\}, and let z \in Z. Consider the equivalence \sigma whose classes are \{\{e,z\}, S \setminus \{e,z\}\}. Now e \sigma z, so that e = e^2 \sigma ez = 0. So e \sigma 0, and since e \neq 0, z = 0. So S = \{e,0\}. Certainly S is a semilattice of order 2.

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